How do you simplify #sqrt(1215)#?

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Answer 1 · 2016-02-22T21:44:12

#9sqrt(15)#

First decompose 1215 in prime factors:

The last digit is 5 so it is multiple o f5:

#1215/5=243#

The sum of the digits of 243 its is 9, therefore it is multiple 3.

#243/3=81#.

We know by the grammar school that #81=9xx9=3^4#.

So #1215=3^5*5#

#Sqrt(1215)=sqrt(3^5*5)=sqrt(3^4*3*5)=3^2sqrt(3*5)=9sqrt(15)#